从数组javascript中删除元素

Question

这是 odin 项目的第四个项目，所有测试都通过了，但是第五个需要删除所有元素的测试失败了，当我运行代码时，它返回一个数组，其中包含原始数组中一半的元素，然后再进行变异。

我不知道为什么在第五次测试中它不返回空数组。

const removeFromArray = function (array, ...deleteElement) {
    for (let i = 0; i < array.length; i++) {
        if (array.includes(deleteElement[i])) {
            array.splice(array.indexOf(deleteElement[i]), 1);
        }
    }
    return array;
};

const randomArray = [1, 2, 3, 4];
console.log(removeFromArray(randomArray, 1, 2, 3, 4));

这是测试

const removeFromArray = require('./removeFromArray')

describe('removeFromArray', () => {
  test('removes a single value', () => {
    expect(removeFromArray([1, 2, 3, 4], 3)).toEqual([1, 2, 4]);
  });
  test('removes multiple values', () => {
    expect(removeFromArray([1, 2, 3, 4], 3, 2)).toEqual([1, 4]);
  });
  test('ignores non present values', () => {
    expect(removeFromArray([1, 2, 3, 4], 7, "tacos")).toEqual([1, 2, 3, 4]);
  });
  test('ignores non present values, but still works', () => {
    expect(removeFromArray([1, 2, 3, 4], 7, 2)).toEqual([1, 3, 4]);
  });
  test.skip('can remove all values', () => {
    expect(removeFromArray([1, 2, 3, 4], 1, 2, 3, 4)).toEqual([]);
  });
  test.skip('works with strings', () => {
    expect(removeFromArray(["hey", 2, 3, "ho"], "hey", 3)).toEqual([2, "ho"]);
  });
  test.skip('only removes same type', () => {
    expect(removeFromArray([1, 2, 3], "1", 3)).toEqual([1, 2]);
  });
});

Answer 1

在 for 中，您正在遍历错误的元素。应该是deleteElement.length，不是array.length

const removeFromArray = function(array, ...deleteElement) {
  for (let i = 0; i < deleteElement.length; i++) {
    if (array.includes(deleteElement[i])) {
      array.splice(array.indexOf(deleteElement[i]), 1);
    }
  }
  return array;
};

const randomArray = [1, 2, 3, 4];
console.log(removeFromArray(randomArray, 1, 2, 3, 4));

编辑： 尽管当数组中有 2 个相同的元素要删除时，这将无法正常工作。这是另一种实现，它删除所有匹配的索引，而不仅仅是第一个。

const removeFromArray = function(array, ...deleteElement) {
  for (let i = 0; i < deleteElement.length; i++) {
    let foundIndex = array.indexOf(deleteElement[i]);
    while (foundIndex !== -1) {
      array.splice(foundIndex, 1);
      foundIndex = array.indexOf(deleteElement[i]);
    }
  }
  return array;
};

const randomArray = [1, 2, 3, 1, 4];
console.log(removeFromArray(randomArray, 1));

Answer 2

你的函数遍历了错误的变量 (array.length)

const removeFromArray = function (array, ...deleteElement) {
    for (let element of deleteElement) {
        if (array.includes(element)) {
            array.splice(array.indexOf(element), 1);
        }
    }
    return array;
};

Answer 3

问题是 for 循环正在动态使用 array.length 属性，这意味着每次通过循环时都会对其进行求值，因此由于您要从原始数组中删除元素，因此该值不是常量并且在某些时候数组长度值和 i 计数器将不同步。

你只需要将数组长度值保存在变量中所以从数组中删除对象后它保持不变。

var arrLen = array.length;
for (let i = 0; i < arrLen; i++) {
    ...
}

更新：

在此处粘贴完整代码以及一个实时示例： https://jsfiddle.net/bcngr/sbk1yhmn/

const removeFromArray = function (array, ...deleteElement) {
    var arrLen = array.length; // Save the length value before any modification to the array
    for (let i = 0; i < arrLen; i++) { // use the variable instead of array.length
        if (array.includes(deleteElement[i])) {
            array.splice(array.indexOf(deleteElement[i]), 1);
        }
    }
    return array;
};

const randomArray = [1, 2, 3, 4];
console.log(removeFromArray(randomArray, 1, 2, 3, 4));